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Let $E$ be a complex Hilbert space. Let $T\in \mathcal{L}(E)$.

It is true that for an arbitrary operator $T\in \mathcal{L}(E)$, we have $$w(T) := \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}=0\Longrightarrow T=0?$$ Or $T$ must be self-adjoint operator?

Thank you.

Alex Ortiz
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Student
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2 Answers2

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Notice that you have the polarization identity

$$\frac{1}{4}\sum_{k=1}^{4} i^{-k}\langle T(u+i^k v), u+i^k v\rangle = \langle Tu, v\rangle$$

Sangchul Lee
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  • Thank you for you answer but I don't understant is the answer is true or false. – Student Dec 22 '17 at 15:27
  • @Student, if $w(T)=0$ then the above equality tells that $\langle Tu,v\rangle=0$ for all $u,v\in E$. From this you can easily deduce that $T=0$. – Sangchul Lee Dec 22 '17 at 15:29
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Notice that $\langle Tu|u\rangle$ is a diagonal matrix element, in the right basis. Is it true a matrix with all diagonal elements zero is the zero matrix? No, for example consider an operator which acts as $\left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)$ on a 2-dimensional subspace and zero on its complement. It satisfies the criterion in the question, but is not the zero matrix.

However it should be true for all diagonalizable matrices, and all normal operators on Hilbert space.

ziggurism
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  • Thank you but $E$ is an infinite-dimensional complex Hilbert space – Student Dec 22 '17 at 14:59
  • @Student that doesn't change the answer. If it fails to be true in finite dimensions, it can only fail harder in infinite dimensions. Infinite dimensions include all the complexities of finite dimensional linear algebra, plus some limit cases. – ziggurism Dec 22 '17 at 14:59
  • Thank you . But why for a normal operator we get the result? – Student Dec 22 '17 at 15:01
  • @Student because a normal operator is diagonalizable. I am thinking of how to prove it without using the spectral theorem. I will edit my answer – ziggurism Dec 22 '17 at 15:03
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    Except that matrix does not satisfy the condition. $T(1,1) = (1,0)$, so $<T(1,1),(1,1)> = 1$. – David C. Ullrich Dec 22 '17 at 15:05
  • @DavidC.Ullrich Well crap. Maybe the claim is true – ziggurism Dec 22 '17 at 15:16
  • The answer just posted by Sangchul Lee shows that it is true for all operators. Whew. I shall delete my answer. – ziggurism Dec 22 '17 at 15:19